NCERT Mathematics (Class 6–10) Cheatsheet

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Number Systems

Class 6–10

Classification of Numbers

Type	Set	Definition	Example
Natural (ℕ)	1, 2, 3, …	Counting numbers (starts at 1)	5, 23, 100
Whole (𝕎)	0, 1, 2, …	Natural numbers + zero	0, 7, 42
Integers (ℤ)	…, −2, −1, 0, 1, …	Whole numbers + negatives	−3, 0, 15
Rational (ℚ)	p/q (q ≠ 0)	Can be expressed as a fraction	1/2, −3/4, 0.5
Irrational	—	Non-repeating, non-terminating decimals	√2, π, e
Real (ℝ)	ℚ ∪ Irrational	All numbers on the number line	−√3, π, 7

The Number Line

Concept	Description
Number Line	A straight horizontal line where each point represents a real number
Direction	Right → increasing; Left → decreasing
Origin	Point 0; integers are equidistant markers
Fractions	Placed between integers by dividing the unit length
Irrationals	√2 ≈ 1.414 is placed approximately between 1 and 2
Negative Numbers	Extends to the left of 0 on the line

LCM & HCF (GCD)

HCF (GCD)Highest Common Factor — largest number that divides all given numbers

LCMLeast Common Multiple — smallest number that is a multiple of all given numbers

RelationHCF(a, b) × LCM(a, b) = a × b (for two numbers)

Prime FactorisationExpress each number as product of primes, then combine

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─── Euclidean Algorithm for HCF ───
Step 1: Divide the larger number by the smaller
Step 2: Replace larger with smaller, smaller with remainder
Step 3: Repeat until remainder = 0
Step 4: Last non-zero remainder is the HCF

Example: HCF(56, 98)
  98 = 56 × 1 + 42
   56 = 42 × 1 + 14
   42 = 14 × 3 + 0   →  HCF = 14

─── LCM by Prime Factorisation ───
12 = 2² × 3
18 = 2 × 3²
LCM = 2² × 3² = 36  (max power of each prime)

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─── Divisibility Rules ───
  By 2  → Last digit is 0, 2, 4, 6, 8
  By 3  → Sum of digits is divisible by 3
  By 4  → Last two digits form a number divisible by 4
  By 5  → Last digit is 0 or 5
  By 6  → Divisible by both 2 and 3
  By 8  → Last three digits form a number divisible by 8
  By 9  → Sum of digits is divisible by 9
  By 11 → (Sum of odd-place digits) − (Sum of even-place digits) is 0 or ±11, ±22, …

Prime & Composite Numbers

Prime NumberA number > 1 that has exactly two factors: 1 and itself

Composite NumberA number > 1 that has more than two factors

1Neither prime nor composite

2Only even prime number; smallest prime

Twin PrimesPairs of primes that differ by 2 (e.g., 3 & 5, 11 & 13)

Sieve of EratosthenesMethod to find all primes ≤ n by eliminating multiples

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─── Prime Factorisation ───
  60 = 2 × 2 × 3 × 5 = 2² × 3¹ × 5¹
  84 = 2² × 3 × 7
  360 = 2³ × 3² × 5

─── Fundamental Theorem of Arithmetic ───
Every composite number can be uniquely expressed (up to order)
as a product of prime factors.

Order of Operations — BODMAS / PEMDAS

Letter	Meaning	Order
B / P	Brackets / Parentheses	1st
O / E	Orders / Exponents (√, xⁿ)	2nd
D / M	Division / Multiplication (left to right)	3rd
M / D	Multiplication / Division (left to right)	3rd
A / A	Addition	4th
S / S	Subtraction	5th

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─── BODMAS Example ───
  24 + 6 × 3 − 18 ÷ (4 + 2)
  = 24 + 6 × 3 − 18 ÷ 6       [Brackets first: 4+2 = 6]
  = 24 + 18 − 3                [Div/Mul: 6×3=18, 18÷6=3]
  = 42 − 3 = 39                [Add/Sub left to right]

Exponents & Power Rules

Rule	Formula	Example
Product (same base)	aᵐ × aⁿ = aᵐ⁺ⁿ	2³ × 2⁴ = 2⁷ = 128
Quotient (same base)	aᵐ ÷ aⁿ = aᵐ⁻ⁿ	5⁶ ÷ 5² = 5⁴ = 625
Power of a power	(aᵐ)ⁿ = aᵐⁿ	(3²)³ = 3⁶ = 729
Power of a product	(ab)ⁿ = aⁿ × bⁿ	(2×3)² = 4 × 9 = 36
Power of a quotient	(a/b)ⁿ = aⁿ/bⁿ	(6/2)³ = 216/8 = 27
Zero exponent	a⁰ = 1 (a ≠ 0)	7⁰ = 1
Negative exponent	a⁻ⁿ = 1/aⁿ	3⁻² = 1/9
Fractional exponent	a^(m/n) = ⁿ√(aᵐ)	8^(2/3) = ³√64 = 4

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─── Squares, Cubes, Square Roots, Cube Roots ───

Squares (n²):  1→1   2→4   3→9   4→16  5→25  6→36
               7→49  8→64  9→81  10→100 11→121 12→144
               13→169 14→196 15→225 20→400 25→625 30→900

Cubes (n³):    1→1   2→8   3→27  4→64  5→125 6→216
               7→343 8→512 9→729 10→1000

Square Root (√):  √4=2  √9=3  √16=4  √25=5  √36=6
                   √49=7  √64=8  √81=9  √100=10  √121=11
                   √144=12  √169=13  √196=14  √225=15

Cube Root (∛):   ∛1=1  ∛8=2  ∛27=3  ∛64=4  ∛125=5
                  ∛216=6  ∛343=7  ∛512=8  ∛729=9  ∛1000=10

Scientific Notation

Standard FormN × 10ⁿ where 1 ≤ N < 10, n is an integer

Large numbers65,000,000 = 6.5 × 10⁷

Small numbers0.00045 = 4.5 × 10⁻⁴

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─── Surds (Irrational Roots) ───

A surd is an irrational number expressed as ⁿ√a
where a is not a perfect nth power.

Simplification examples:
  √50  = √(25 × 2)  = 5√2
  √72  = √(36 × 2)  = 6√2
  √98  = √(49 × 2)  = 7√2
  √128 = √(64 × 2)  = 8√2
  √200 = √(100 × 2) = 10√2

Operations on surds:
  a√m + b√m  = (a + b)√m       [Like surds]
  a√m × b√n  = ab√(mn)
  a√m ÷ b√n  = (a/b)√(m/n)
  (√m)²      = m

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─── Laws of Logarithms (Class 9 — basic intro) ───

If aᵐ = N, then logₐ N = m  (a > 0, a ≠ 1, N > 0)

Product Rule:     logₐ(M × N)  = logₐ M + logₐ N
Quotient Rule:    logₐ(M / N)  = logₐ M − logₐ N
Power Rule:       logₐ Mⁿ      = n × logₐ M
Base Change:      logₐ N       = log N / log a
Reciprocal:       logₐ (1/N)   = −logₐ N

Special Values:
  log₁₀ 1    = 0
  log₁₀ 10   = 1
  logₐ a     = 1
  logₐ 1     = 0

Common Logarithm: log₁₀ (base 10, written as log)
Natural Logarithm: logₑ = ln (base e ≈ 2.718)

Number System Conversions (Binary ↔ Decimal)

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─── Decimal to Binary ───
  Divide by 2, record remainders bottom-up

  25 → Decimal to Binary:
  25 ÷ 2 = 12 R 1
  12 ÷ 2 = 6  R 0
   6 ÷ 2 = 3  R 0
   3 ÷ 2 = 1  R 1
   1 ÷ 2 = 0  R 1
  Read bottom-up: 25₁₀ = 11001₂

─── Binary to Decimal ───
  Multiply each digit by 2^position (from right, 0-indexed)

  11001₂:
  1×2⁴ + 1×2³ + 0×2² + 0×2¹ + 1×2⁰
  = 16 + 8 + 0 + 0 + 1 = 25₁₀

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─── Properties of Rational Numbers ───

Closure Property:
  a + b, a − b, a × b ∈ ℚ (for a, b ∈ ℚ)
  a ÷ b ∈ ℚ only when b ≠ 0

Commutativity:     a + b = b + a;   a × b = b × a
Associativity:     (a+b)+c = a+(b+c); (a×b)×c = a×(b×c)
Identity:          a + 0 = a;   a × 1 = a
Inverse:           a + (−a) = 0;   a × (1/a) = 1
Distributivity:    a × (b + c) = a × b + a × c

─── Real Numbers — Dedekind Cut (concept) ───
Every real number divides rational numbers into two
non-empty sets L (lower) and U (upper) such that:
  • Every element of L < every element of U
  • L has no greatest element
This is the Dedekind cut definition of real numbers.

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Exam Tip: For HCF/LCM word problems, remember: HCF is used when dividing things into equal parts (e.g., cutting ropes), and LCM is used when finding when events happen together (e.g., bells ringing together).

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Algebra

Class 6–10

Algebra Fundamentals

Term	Definition	Example
Variable	A symbol that represents an unknown value	x, y, n
Constant	A fixed value that does not change	5, −3, π
Term	Variable, constant, or product of both	3x, 7, 2xy²
Coefficient	Numerical factor of a term	In 5x²y, coeff of x² is 5y
Expression	Combination of terms using + / −	3x + 2y − 7
Equation	Expression with an = sign	2x + 3 = 11
Like Terms	Same variables raised to same powers	3x² and −5x²
Unlike Terms	Different variables or powers	3x² and 5x³

Polynomials — Classification

Degree	Name	Example
0	Constant	7
1	Linear	2x + 3
2	Quadratic	x² − 5x + 6
3	Cubic	x³ + 2x² − x + 1
4	Quartic / Biquadratic	x⁴ − 16
n	n-th degree	aₙxⁿ + … + a₁x + a₀

Standard FormP(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀

Zero of PolynomialValue of x for which P(x) = 0

Zeroes of Linearax + b = 0 → x = −b/a

Linear Factor TheoremIf P(a) = 0, then (x − a) is a factor of P(x)

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─── Operations on Polynomials ───

Addition:  Add like terms
  (2x² + 3x − 1) + (x² − 2x + 5) = 3x² + x + 4

Subtraction: Subtract like terms
  (2x² + 3x − 1) − (x² − 2x + 5) = x² + 5x − 6

Multiplication: Multiply each term
  (x + 2)(x + 3) = x² + 3x + 2x + 6 = x² + 5x + 6

Division: Long division or synthetic division
  (x² + 5x + 6) ÷ (x + 2) = x + 3  (since x²+5x+6 = (x+2)(x+3))

Factorization Methods

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─── Method 1: Common Factor ───
  6x² + 9x = 3x(2x + 3)

─── Method 2: Factorising by Grouping ───
  x² + 5x + 6 = x² + 2x + 3x + 6
              = x(x + 2) + 3(x + 2)
              = (x + 2)(x + 3)

─── Method 3: Split the Middle Term ───
  x² + 7x + 12
  Product = 12,  Sum = 7  →  factors: 3, 4
  = x² + 3x + 4x + 12
  = x(x + 3) + 4(x + 3)
  = (x + 3)(x + 4)

─── Method 4: Difference of Squares ───
  a² − b² = (a + b)(a − b)
  4x² − 9  = (2x)² − 3² = (2x + 3)(2x − 3)

─── Method 5: Sum/Difference of Cubes ───
  a³ + b³  = (a + b)(a² − ab + b²)
  a³ − b³  = (a − b)(a² + ab + b²)

Algebraic Identities

Identity	Formula
Square of Sum	(a + b)² = a² + 2ab + b²
Square of Difference	(a − b)² = a² − 2ab + b²
Product of Sum & Diff	(a + b)(a − b) = a² − b²
Cube of Sum	(a + b)³ = a³ + 3a²b + 3ab² + b³
Cube of Difference	(a − b)³ = a³ − 3a²b + 3ab² − b³
Sum of Cubes	a³ + b³ = (a + b)(a² − ab + b²)
Difference of Cubes	a³ − b³ = (a − b)(a² + ab + b²)
Square of Triple	(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
a² + b²	a² + b² = (a + b)² − 2ab = (a − b)² + 2ab
x³ + y³ + z³ − 3xyz	(x + y + z)(x² + y² + z² − xy − yz − zx)

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Remember: If x + y + z = 0, then x³ + y³ + z³ = 3xyz. This is a very useful shortcut in exams!

Linear Equations — One Variable

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─── Solving ax + b = 0 ───
  Rule: Whatever you do to one side, do to the other.

  Example: Solve 3x + 7 = 22
  3x = 22 − 7 = 15
  x = 15 / 3 = 5

  Example: Solve (2x − 3)/5 = (x + 1)/3
  Cross multiply: 3(2x − 3) = 5(x + 1)
  6x − 9 = 5x + 5
  6x − 5x = 5 + 9
  x = 14

Linear Equations — Two Variables

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─── Graphical Solution ───
  a₁x + b₁y = c₁    …(i)
  a₂x + b₂y = c₂    …(ii)

  Plot both lines on the graph.
  The point of intersection (x, y) is the solution.

─── Number of Solutions ───
  a₁/a₂ ≠ b₁/b₂  → Unique solution (lines intersect)   → Consistent
  a₁/a₂ = b₁/b₂ ≠ c₁/c₂ → No solution (parallel lines) → Inconsistent
  a₁/a₂ = b₁/b₂ = c₁/c₂ → Infinitely many (coincident) → Consistent

─── Substitution Method ───
  From eq(i): x = (c₁ − b₁y) / a₁
  Substitute into eq(ii), solve for y, then find x.

─── Elimination Method ───
  Multiply equations to make coefficients equal.
  Add/subtract to eliminate one variable.

─── Cross-Multiplication Method ───
         x         y          1
  ───────────── = ────────── = ─────────────
   b₁c₂−b₂c₁   c₁a₂−c₂a₁   a₁b₂−a₂b₁

Quadratic Equations (Class 10)

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─── Standard Form ───
  ax² + bx + c = 0  (a ≠ 0)

─── Solving by Factorisation ───
  x² − 5x + 6 = 0
  x² − 2x − 3x + 6 = 0
  x(x − 2) − 3(x − 2) = 0
  (x − 2)(x − 3) = 0
  x = 2  or  x = 3

─── Quadratic Formula ───
           −b ± √(b² − 4ac)
  x = ─────────────────────
              2a

─── Discriminant (D) = b² − 4ac ───
  D > 0  → Two distinct real roots
  D = 0  → Two equal real roots (one repeated root)
  D < 0  → No real roots (complex roots)

─── Sum & Product of Roots ───
  If α, β are roots of ax² + bx + c = 0:
    α + β = −b/a
    α × β = c/a

─── Nature of Roots ───
  D > 0 & perfect square → roots are rational & distinct
  D > 0 & not square     → roots are irrational & distinct
  D = 0                  → roots are equal & rational

Arithmetic Progression — AP (Class 10)

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─── Arithmetic Progression ───
  A sequence where each term differs from the previous by
  a constant called the Common Difference (d).

  General Form: a, a+d, a+2d, a+3d, …
  First term = a,  Common difference = d

─── nth Term (General Term) ───
  aₙ = a + (n − 1)d

─── Sum of First n Terms ───
  Sₙ = n/2 × [2a + (n − 1)d]
     = n/2 × [a + aₙ]    (using last term)

─── Key Properties ───
  • If a, b, c are in AP → 2b = a + c
  • aₙ − aₘ = (n − m)d
  • Sum of n terms can also be found by pairing:
    Sₙ = n/2 × (first term + last term)

─── Example ───
  AP: 3, 7, 11, 15, …
  a = 3, d = 4
  a₂₀ = 3 + 19 × 4 = 79
  S₂₀ = 20/2 × (2×3 + 19×4) = 10 × 82 = 820

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─── Geometric Progression — GP (Class 10 Intro) ───
  A sequence where each term is obtained by multiplying
  the previous term by a constant called the Common Ratio (r).

  General Form: a, ar, ar², ar³, …
  First term = a,  Common ratio = r

  nth Term:   aₙ = a × r^(n−1)

  Sum of n terms (r ≠ 1):
              a(rⁿ − 1)
  Sₙ = ─────────────
              r − 1

  Key Property:  If a, b, c are in GP → b² = a × c

  Example: 2, 6, 18, 54, …
  a = 2, r = 3
  a₅ = 2 × 3⁴ = 162
  S₅ = 2(3⁵−1)/(3−1) = 2 × 242/2 = 242

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Common Mistake: When using the quadratic formula, be careful with signs. Write x = (−b ± √D) / 2a. Also verify your roots by substituting back into the original equation.

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Geometry

Class 6–10

Basic Geometric Terms

Term	Definition
Point	A dot that has position but no size; represented by a capital letter (A, B, P)
Line	A collection of points extending infinitely in both directions; denoted as AB or ↔
Line Segment	Part of a line with two endpoints; denoted as AB or ↔
Ray	Part of a line with one endpoint, extending infinitely in one direction
Plane	A flat surface extending infinitely in all directions
Angle	Figure formed by two rays with a common endpoint (vertex)

Types of Angles

Type	Measure	Description
Acute Angle	0° < θ < 90°	Less than a right angle
Right Angle	θ = 90°	Exactly one-quarter turn
Obtuse Angle	90° < θ < 180°	Between right and straight
Straight Angle	θ = 180°	Half turn; forms a straight line
Reflex Angle	180° < θ < 360°	Greater than a straight angle
Complete Angle	θ = 360°	Full turn
Complementary	θ₁ + θ₂ = 90°	Two angles that add to 90°
Supplementary	θ₁ + θ₂ = 180°	Two angles that add to 180°
Adjacent	Share a common arm & vertex	No overlap of interiors
Linear Pair	Adjacent + supplementary	Non-common arms form a line
Vertically Opposite	Opposite angles at intersection	Always equal

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─── Angle Relationships ───

Vertically Opposite Angles:
  When two lines intersect:
    ∠1 = ∠3  and  ∠2 = ∠4  (vertically opposite)

Linear Pair Axiom:
  If a ray stands on a line → adjacent angles sum to 180°
  ∠AOB + ∠BOC = 180°  (if OB stands on line AC)

Parallel Lines + Transversal:
  Corresponding angles  (F-shape)  → Equal
  Alternate interior     (Z-shape)  → Equal
  Alternate exterior                → Equal
  Co-interior (same side)           → Supplementary (sum = 180°)
  Vertically opposite                → Equal

Triangles — Classification

By Sides	Description	By Angles	Description
Equilateral	All sides equal	Acute	All angles < 90°
Isosceles	Two sides equal	Right	One angle = 90°
Scalene	All sides different	Obtuse	One angle > 90°
		Equiangular	All angles = 60°

Angle Sum Property∠A + ∠B + ∠C = 180° (for any triangle)

Exterior AngleEquals sum of two interior opposite angles

Triangle InequalitySum of any two sides > third side

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─── Congruence of Triangles ───
  Two triangles are congruent if they have the same
  shape and size (all sides & angles match).

Criteria:
  SSS  — Side-Side-Side
         All three sides are equal

  SAS  — Side-Angle-Side
         Two sides and included angle are equal

  ASA  — Angle-Side-Angle
         Two angles and included side are equal

  AAS  — Angle-Angle-Side
         Two angles and a non-included side are equal

  RHS  — Right angle-Hypotenuse-Side
         Right angle, hypotenuse, and one side equal
         (for right-angled triangles only)

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─── Similarity of Triangles ───
  Two triangles are similar if they have the same shape
  (angles equal; sides proportional).

Criteria:
  AA   — Angle-Angle
         Any two angles are equal (third is automatically equal)

  SSS  — Side-Side-Side (proportional)
         All sides in the same ratio: AB/DE = BC/EF = CA/FD

  SAS  — Side-Angle-Side (proportional)
         Two sides in same ratio, included angle equal

Basic Proportionality Theorem (Thales):
  If a line is parallel to one side of a triangle and
  intersects the other two sides, it divides them in
  the same ratio.

  AD/DB = AE/EC  (if DE ∥ BC in △ABC)

Converse:
  If AD/DB = AE/EC, then DE ∥ BC

Pythagoras Theorem

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─── Pythagoras Theorem (Class 10) ───
  In a right-angled triangle:

        hypotenuse² = perpendicular² + base²

        AC² = AB² + BC²

  Common Pythagorean Triplets:
  (3, 4, 5)  → 3² + 4² = 9 + 16 = 25 = 5²
  (5, 12, 13) → 5² + 12² = 25 + 144 = 169 = 13²
  (8, 15, 17) → 64 + 225 = 289 = 17²
  (7, 24, 25) → 49 + 576 = 625 = 25²
  (6, 8, 10)  → 36 + 64 = 100 = 10²  (multiple of 3,4,5)

─── Converse of Pythagoras Theorem ───
  If in △ABC, AC² = AB² + BC²,
  then ∠B = 90° (right angle at B).

─── Applications ───
  • Finding distance between two points
  • Checking if a triangle is right-angled
  • Finding unknown sides in right triangles

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Exam Tip: Memorize the Pythagorean triplets (3,4,5), (5,12,13), (8,15,17), and (7,24,25). In MCQs, you can quickly check for right triangles using these.

Quadrilaterals

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─── Angle Sum Property ───
  Sum of interior angles of any quadrilateral = 360°

─── Parallelogram Properties ───
  • Opposite sides are equal and parallel
  • Opposite angles are equal
  • Diagonals bisect each other
  • Each diagonal divides it into two congruent triangles

─── Rectangle ───
  • All parallelogram properties
  • All angles = 90°
  • Diagonals are equal in length

─── Rhombus ───
  • All parallelogram properties
  • All sides are equal
  • Diagonals bisect at 90° (perpendicular)
  • Diagonals bisect the angles

─── Square ───
  • All rectangle properties + all rhombus properties
  • All sides equal, all angles = 90°
  • Diagonals are equal AND perpendicular

─── Kite ───
  • Two pairs of adjacent sides equal
  • Diagonals are perpendicular
  • One diagonal bisects the other
  • One pair of opposite angles equal

Circles

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─── Key Terms ───
  Radius (r):     Centre to any point on circle
  Diameter (d):   Twice the radius; d = 2r
  Chord:          Line segment joining two points on circle
  Diameter is the longest chord

─── Tangent Properties ───
  • A tangent is perpendicular to the radius at the point of contact
  • Tangents from an external point to a circle are equal in length
  • Length of tangent² = PA × PB  (if PAB is a secant)

─── Angle Subtended by an Arc ───
  • Angle subtended by an arc at the centre = 2 × angle at
    any point on the remaining part of the circle
  ∠AOB = 2 × ∠ACB  (O = centre, C on circle)

─── Cyclic Quadrilateral ───
  A quadrilateral whose all vertices lie on a circle.
  • Opposite angles are supplementary: ∠A + ∠C = 180°
  • ∠B + ∠D = 180°
  • Exterior angle = interior opposite angle

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─── Important Constructions ───

1. Angle Bisector:
   Using compass, draw arcs from vertex. Draw arcs from
   intersection points. Join vertex to new intersection.

2. Perpendicular Bisector:
   From both endpoints, draw arcs above and below the line.
   Join the two intersection points.

3. Triangle given SSS:
   Draw one side, then use compass with the other two
   sides' lengths to locate the third vertex.

4. Triangle given SAS:
   Draw one side, construct the angle at one end, then
   mark the other side's length along the angle ray.

5. Triangle given ASA:
   Draw one side, construct angles at both ends. The angle
   rays intersect at the third vertex.

6. Tangent from an External Point:
   Join external point P to centre O.
   Find midpoint M of PO.
   Draw circle with centre M, radius MO.
   This circle intersects the original circle at T₁, T₂.
   PT₁ and PT₂ are the required tangents.

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Mensuration

Class 6–10

2D — Perimeter & Area Formulas

Shape	Perimeter (P)	Area (A)
Triangle	a + b + c	½ × base × height
Equilateral △	3a	(√3/4) × a²
Right △	a + b + c	½ × base × height = ½ × a × b
Rectangle	2(l + b)	l × b
Square	4a	a²
Parallelogram	2(a + b)	base × height
Rhombus	4a	½ × d₁ × d₂
Trapezium	a + b + c + d	½ × (a + b) × h (parallel sides a, b)
Circle	2πr (or πd)	πr²
Semi-circle	πr + 2r	½πr²
Ring (Annulus)	—	π(R² − r²)

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─── Heron's Formula (Class 9) ───
  For a triangle with sides a, b, c:

  Step 1: Find semi-perimeter
    s = (a + b + c) / 2

  Step 2: Apply Heron's formula
    Area = √[s(s − a)(s − b)(s − c)]

  Example:  a = 5, b = 6, c = 7
    s = (5 + 6 + 7) / 2 = 9
    A = √[9 × (9−5) × (9−6) × (9−7)]
    A = √[9 × 4 × 3 × 2]
    A = √216 = 6√6 ≈ 14.70 sq units

Text

─── Sector of a Circle (Class 10) ───

  Length of Arc:
    l = (θ/360°) × 2πr    (θ in degrees)
    l = r × θ              (θ in radians)

  Area of Sector:
    A = (θ/360°) × πr²    (θ in degrees)

  Area of Segment:
    = Area of sector − Area of triangle
    = (θ/360°) × πr² − ½r²sinθ

  Perimeter of Sector:
    = 2r + l = 2r + (θ/360°) × 2πr

3D — Surface Area & Volume Formulas

Shape	Total Surface Area	Volume
Cube (side a)	6a²	a³
Cuboid (l, b, h)	2(lb + bh + hl)	l × b × h
Cylinder (r, h)	2πr(r + h)	πr²h
Hollow Cylinder	2π(R + r)(R − r + h)	π(R² − r²)h
Cone (r, h, l)	πr(r + l)	⅓πr²h
Sphere (r)	4πr²	(4/3)πr³
Hemisphere	3πr² (curved + base)	(2/3)πr³
Hollow Sphere	4π(R² + r²)	(4/3)π(R³ − r³)

Text

─── Key Relations ───

Cone:
  Slant height:  l = √(r² + h²)

Sphere:
  Diameter = 2r

─── Composite Solids ───
  Volume of a solid = Sum of volumes of its parts

  Example: A toy is a cone mounted on a hemisphere
  Total Volume = Volume of Cone + Volume of Hemisphere
               = ⅓πr²h + ⅔πr³

─── Conversion of Solids (Melt & Recast) ───
  When one solid is melted and recast into another,
  the volume remains constant.

  Volume₁ = Volume₂

  Example: A metallic sphere of r = 6 cm is melted
  and recast into a cylinder of h = 8 cm. Find r.

  (4/3)π(6)³ = πr²(8)
  (4/3)(216) = 8r²
  288 = 8r²
  r² = 36 → r = 6 cm

Text

─── Types of Surface Area ───

Lateral/Curved Surface Area (CSA):
  • Cube:  4a²
  • Cuboid: 2h(l + b)
  • Cylinder: 2πrh
  • Cone: πrl
  • Sphere: 4πr²
  • Hemisphere: 2πr²

Total Surface Area (TSA) = CSA + Area of Bases
  • Cylinder: 2πrh + 2πr² = 2πr(r + h)
  • Cone: πrl + πr² = πr(l + r)
  • Hemisphere: 2πr² + πr² = 3πr²

⚠️

Common Mistake: For cylinder problems, read carefully — "curved surface area" means 2πrh while "total surface area" means 2πr(r + h). Similarly for cones: CSA = πrl, TSA = πr(l + r).

💡

Exam Tip: In melt-and-recast problems, volume is always conserved. In surface area calculations, total area changes (usually increases). Always check whether the question asks for CSA or TSA.

📊

Statistics & Probability

Class 6–10

Data Collection & Organisation

Term	Definition
Raw Data	Unorganised collection of numbers/observations
Frequency	Number of times a value occurs in the data
Tally Marks	Quick counting method: \|\|\|\| grouped as \|\|\|\| for 5
Class Interval	Range of data values grouped together (e.g., 10–20)
Class Width	Difference between upper and lower limits of a class
Class Mark	Midpoint of a class = (upper + lower) / 2
Range	Maximum value − Minimum value
Cumulative Frequency	Running total of frequencies up to that class

Measures of Central Tendency

Text

─── Mean (Average) ───

Method 1 — Direct Mean (Ungrouped):
        Σxᵢ
  Mean = ─────
         n

Method 2 — Direct Mean (Grouped):
           Σ(fᵢ × xᵢ)
  Mean = ─────────────
           Σfᵢ

Method 3 — Assumed Mean Method:
           Σfᵢdᵢ
  Mean = A + ────────
            Σfᵢ
  where dᵢ = xᵢ − A  (A = assumed mean)

Method 4 — Step Deviation Method:
           Σfᵢuᵢ
  Mean = A + h × ────────
                 Σfᵢ
  where uᵢ = (xᵢ − A) / h  (h = class width)

Median & Mode

Text

─── Median (Ungrouped) ───
  Step 1: Arrange data in ascending order
  Step 2:
    If n is odd:  Median = Value at position (n+1)/2
    If n is even: Median = Average of values at n/2 and (n/2)+1

─── Median (Grouped / Continuous) ───
         n
  Find ─── , locate the cumulative frequency just greater
         2
  than this value. The corresponding class is the median class.

          n
         ─── − cf
          2
  Median = l + ────────── × h
            f

  l  = lower limit of median class
  n  = total frequency
  cf = cumulative frequency of class before median class
  f  = frequency of median class
  h  = class width

─── Mode (Grouped) ───
  Modal class = class with highest frequency

           f₁ − f₀
  Mode = l + ────────────── × h
           2f₁ − f₀ − f₂

  l  = lower limit of modal class
  f₁ = frequency of modal class
  f₀ = frequency of class before modal class
  f₂ = frequency of class after modal class
  h  = class width

─── Empirical Relationship ───
  3 × Median = Mode + 2 × Mean

💡

Remember: The empirical relationship 3 Median = Mode + 2 Mean (or Mode = 3 Median − 2 Mean) is very useful when you need to find one measure when the other two are known.

Graphical Representation of Data

Graph Type	Description	Used For
Bar Graph	Bars of equal width, heights proportional to frequency	Comparing discrete categories
Histogram	Connected bars (no gaps), area proportional to frequency	Continuous/frequency distribution
Frequency Polygon	Points at class midpoints connected by lines	Comparing two distributions
Ogive	Cumulative frequency curve (less than / more than type)	Finding median graphically
Pie Chart	Circle divided into sectors proportional to values	Showing parts of a whole

Text

─── Ogive (Cumulative Frequency Curve) ───

Step 1: Calculate cumulative frequencies (less than type)
Step 2: Plot points (upper limit, cumulative frequency)
Step 3: Join points with a smooth free-hand curve

Median from Ogive:
  Draw a horizontal line from n/2 on the y-axis.
  Where it meets the ogive, drop a perpendicular to x-axis.
  That x-value is the median.

─── Pie Chart ───
  Angle of a sector = (value / total) × 360°

  Example: If 40% of students play cricket
  Angle = (40/100) × 360° = 144°

Probability (Class 10)

Text

─── Basic Definition ───

  P(E) =  Number of favourable outcomes
         ─────────────────────────────────
           Total number of possible outcomes

  0 ≤ P(E) ≤ 1

  P(E) = 0  → Impossible event
  P(E) = 1  → Certain (sure) event

─── Key Rules ───

  Complementary Events:
    P(E) + P(not E) = 1
    P(not E) = 1 − P(E)

  Sum of probabilities of all outcomes = 1

─── Sample Space (S) ───
  The set of all possible outcomes of an experiment.

─── Coin Problems ───
  One coin:    S = {H, T}                  n(S) = 2
  Two coins:   S = {HH, HT, TH, TT}       n(S) = 4
  Three coins: S = {HHH, HHT, HTH, HTT,
                     THH, THT, TTH, TTT}    n(S) = 8

─── Dice Problems ───
  One die:     S = {1, 2, 3, 4, 5, 6}      n(S) = 6
  Two dice:    n(S) = 36  (6 × 6 outcomes)

  P(sum = 7)  = 6/36 = 1/6
  P(sum = 12) = 1/36
  P(doublet)  = 6/36 = 1/6

─── Card Problems ───
  Total cards = 52
  4 suits (13 each): Hearts ♥, Diamonds ♦, Clubs ♣, Spades ♠
  Red cards: 26    Black cards: 26
  Face cards: 12 (J, Q, K of each suit)
  Aces: 4

  P(King)     = 4/52 = 1/13
  P(Red Ace)  = 2/52 = 1/26
  P(not King) = 48/52 = 12/13

⚠️

Exam Tip: For probability, always list out the complete sample space first. For two dice, draw the 6×6 table to avoid counting errors. Remember: "or" means addition (+), "and" means multiplication (×) for independent events.

📐

Trigonometry

Class 10

Trigonometric Ratios

Text

─── In a Right-Angled Triangle ABC (right angle at B) ───

        C
       /|
      / |
   h /  | p    h = hypotenuse (AC)
    /   |       p = perpendicular (BC — opposite to angle)
   /θ___|       b = base (AB — adjacent to angle)
   A     B

  sin θ  = p / h  = BC / AC  = Opposite / Hypotenuse
  cos θ  = b / h  = AB / AC  = Adjacent / Hypotenuse
  tan θ  = p / b  = BC / AB  = Opposite / Adjacent

  cosec θ = h / p = AC / BC = 1 / sin θ
  sec θ   = h / b = AC / AB = 1 / cos θ
  cot θ   = b / p = AB / BC = 1 / tan θ

─── Mnemonic ───
  "Some People Have Curly Brown Hair Through Proper Brushing"
  Sin = P/H,  Cos = B/H,  Tan = P/B
  Cosec = H/P,  Sec = H/B,  Cot = B/P

Trigonometric Ratios — Standard Angles

Angle	sin θ	cos θ	tan θ	cosec θ	sec θ	cot θ
0°	0	1	0	∞	1	∞
30°	1/2	√3/2	1/√3	2	2/√3	√3
45°	1/√2	1/√2	1	√2	√2	1
60°	√3/2	1/2	√3	2/√3	2	1/√3
90°	1	0	∞	1	∞	0

💡

Memory Trick: For the table above, remember the sequence for sin θ: 0, 1/2, 1/√2, √3/2, 1. Then cos θ is the reverse. Tan θ = sin/cos. The reciprocals give cosec, sec, cot.

Trigonometric Identities (Class 10)

Text

─── Fundamental Identities ───

  (1)  sin²θ + cos²θ  = 1
  (2)  1 + tan²θ      = sec²θ
  (3)  1 + cot²θ      = cosec²θ

─── Derived Forms ───

  sin²θ  = 1 − cos²θ         cos²θ  = 1 − sin²θ
  tan²θ  = sec²θ − 1         cot²θ  = cosec²θ − 1
  sin θ  = √(1 − cos²θ)      cos θ  = √(1 − sin²θ)

─── Using Identities — Steps ───
  1. Start from the more complex side
  2. Express everything in sin and cos
  3. Use sin²θ + cos²θ = 1
  4. Simplify

─── Example: Prove (1 + tan²A) / (1 + cot²A) = tan²A

  LHS = (sec²A) / (cosec²A)
      = (1/cos²A) / (1/sin²A)
      = sin²A / cos²A
      = tan²A  ✓

Text

─── Complementary Angle Ratios ───

  Two angles are complementary if they add up to 90°.

  sin(90° − θ)  = cos θ          cos(90° − θ)  = sin θ
  tan(90° − θ)  = cot θ          cot(90° − θ)  = tan θ
  sec(90° − θ)  = cosec θ        cosec(90° − θ)= sec θ

─── Examples ───
  sin(90° − 30°) = cos 30° = √3/2
  tan(90° − 45°) = cot 45° = 1
  cosec(90° − 60°) = sec 60° = 2

Heights & Distances — Applications

Text

─── Angle of Elevation & Depression ───

  Angle of Elevation:
    The angle formed by the line of sight (looking up)
    with the horizontal.

  Angle of Depression:
    The angle formed by the line of sight (looking down)
    with the horizontal.

─── Key Points ───
  • Angle of elevation from A = Angle of depression from B
    (when A and B are at the same horizontal level)
  • The line of sight is the hypotenuse
  • The horizontal is always the adjacent side

─── Typical Problem Setup ───

  Tower/building problem:
          Tower
          |  |
          |  | h = ?
          |  |
          |  |
    θ_____|  |
    |  d  |  |
    ˉˉˉˉˉˉˉˉˉˉ  (ground)

    tan θ = h / d
    h = d × tan θ

─── Example ───
  From a point 30 m away from the foot of a tower,
  the angle of elevation is 30°. Find the height.

  tan 30° = h / 30
  1/√3 = h / 30
  h = 30/√3 = 10√3 ≈ 17.32 m

⚠️

Common Mistake: Drawing the angle of depression from the wrong point. Always draw it from the observer's eye level horizontally. Also, for problems involving a tower and a building, don't forget to add/subtract the observer's height.

📍

Coordinate Geometry

Class 9–10

Coordinate Plane Basics

Text

─── Cartesian Coordinate System ───

  • Two perpendicular lines: X-axis (horizontal) and Y-axis (vertical)
  • Their intersection is the Origin O(0, 0)
  • Every point is represented as P(x, y)
  • Four quadrants:
    Q1: (+, +)   Q2: (−, +)
    Q3: (−, −)   Q4: (+, −)

  • Distance of point from X-axis = |y|
  • Distance of point from Y-axis = |x|

Distance Formula

Text

─── Distance between two points ───

  P(x₁, y₁) and Q(x₂, y₂):

            ─────────────────────
  d(P, Q) = √(x₂ − x₁)² + (y₂ − y₁)²

─── Special Cases ───
  • Distance from origin:
    d = √(x² + y²)

  • Horizontal distance (same y):
    d = |x₂ − x₁|

  • Vertical distance (same x):
    d = |y₂ − y₁|

─── Collinearity Check ───
  Three points A, B, C are collinear if:
    AB + BC = AC
  (i.e., sum of two shorter distances = longest distance)

─── Example ───
  A(1, 2) and B(4, 6):
  d = √((4−1)² + (6−2)²) = √(9 + 16) = √25 = 5

Section Formula

Text

─── Internal Division ───
  Point P divides A(x₁, y₁) and B(x₂, y₂) in ratio m:n

         mx₂ + nx₁        my₂ + ny₁
    x = ───────────  ,  y = ───────────
          m + n             m + n

─── Midpoint Formula (m = n = 1) ───
         x₁ + x₂        y₁ + y₂
    x = ─────────  ,  y = ─────────
            2               2

─── External Division ───
  Point P divides AB externally in ratio m:n

         mx₂ − nx₁        my₂ − ny₁
    x = ───────────  ,  y = ───────────
          m − n             m − n

─── Centroid of Triangle ───
  The point where the three medians meet.

  If vertices are A(x₁, y₁), B(x₂, y₂), C(x₃, y₃):

         x₁ + x₂ + x₃        y₁ + y₂ + y₃
    Gx = ────────────   ,  Gy = ────────────
              3                   3

Text

─── Area of Triangle (Coordinate Method) ───

  For vertices A(x₁, y₁), B(x₂, y₂), C(x₃, y₃):

                    1
  Area = ───── × | x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂) |
                  2

─── Determinant Form ───
         | x₁  y₁  1 |
  A = ½ | x₂  y₂  1 |
         | x₃  y₃  1 |

  (expand by adding products of diagonals)

─── Collinearity Check (Alternative) ───
  Three points are collinear if the area of the
  triangle formed by them is zero.

─── Example ───
  A(0, 0), B(4, 0), C(0, 3):
  Area = ½ |0(0−3) + 4(3−0) + 0(0−0)|
       = ½ |0 + 12 + 0|
       = 6 sq units

Equation of a Line (Class 10)

Text

─── Various Forms ───

1. Slope-Intercept Form:
   y = mx + c
   m = slope,  c = y-intercept
   Slope = (y₂ − y₁) / (x₂ − x₁)

2. Point-Slope Form:
   y − y₁ = m(x − x₁)
   Passes through (x₁, y₁) with slope m

3. Two-Point Form:
     y₂ − y₁
   y − y₁ = ────────── × (x − x₁)
     x₂ − x₁
   Passes through (x₁, y₁) and (x₂, y₂)

4. Slope of Parallel Lines:
   m₁ = m₂  (slopes are equal)

5. Slope of Perpendicular Lines:
   m₁ × m₂ = −1

6. General Form:
   ax + by + c = 0
   Slope m = −a/b
   y-intercept = −c/b
   x-intercept = −c/a

─── Conditions for Lines ───
  For lines a₁x + b₁y + c₁ = 0  and  a₂x + b₂y + c₂ = 0:

  Intersecting:   a₁/a₂ ≠ b₁/b₂
  Parallel:       a₁/a₂ = b₁/b₂ ≠ c₁/c₂
  Coincident:     a₁/a₂ = b₁/b₂ = c₁/c₂

💡

Exam Tip: For coordinate geometry, always draw a rough diagram. For area problems, use the formula in determinant form — it's easier to remember. If the area comes out negative, take the absolute value.

📋

Formula Sheet — By Class

Quick Revision

Class 6 — Key Formulas

Topic	Formula
Perimeter of Rectangle	P = 2(l + b)
Area of Rectangle	A = l × b
Perimeter of Square	P = 4 × side
Area of Square	A = side²
Perimeter of Triangle	P = a + b + c
Area of Triangle	A = ½ × base × height
Perimeter of Equilateral △	P = 3 × side
LHS = RHS	Left Hand Side = Right Hand Side
Simple Interest	I = (P × R × T) / 100
Amount	A = P + I
Whole numbers	𝕎 = {0, 1, 2, 3, …}
Integers	ℤ = {…, −2, −1, 0, 1, 2, …}

Class 7 — Key Formulas

Topic	Formula
Area of Parallelogram	A = base × height
Area of Triangle	A = ½ × base × height
Area of Circle	A = πr²
Circumference of Circle	C = 2πr = πd
Area of Path (outer)	π(R² − r²)
Simple Interest	I = (P × R × T) / 100
Compound Interest	A = P(1 + R/100)ⁿ
Profit %	(Profit / CP) × 100
Loss %	(Loss / CP) × 100
SP at Profit	SP = CP × (1 + Profit%/100)
SP at Loss	SP = CP × (1 − Loss%/100)
Rational Number	p/q where q ≠ 0
Laws of Exponents	aᵐ × aⁿ = aᵐ⁺ⁿ; (aᵐ)ⁿ = aᵐⁿ

Class 8 — Key Formulas

Topic	Formula
Square	Area = a²; Diagonal = a√2
Cube	TSA = 6a²; Volume = a³; Diagonal = a√3
Cuboid	TSA = 2(lb + bh + hl); V = lbh
Cylinder (CSA)	CSA = 2πrh
Cylinder (TSA)	TSA = 2πr(r + h)
Cylinder (Volume)	V = πr²h
Area of Trapezium	A = ½ × (a + b) × h
Area of Rhombus	A = ½ × d₁ × d₂
Compound Interest	A = P(1 + R/100)ⁿ
CI half-yearly	A = P(1 + R/200)²ⁿ
Pythagorean Theorem	H² = P² + B²
Algebraic Identity	(a + b)² = a² + 2ab + b²
Algebraic Identity	(a − b)² = a² − 2ab + b²
Algebraic Identity	(a + b)(a − b) = a² − b²
Exponent Law	a⁰ = 1; a⁻ⁿ = 1/aⁿ
Squares 1-30	1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289,324,361,400,441,484,529,576,625,676,729,784,841,900

Class 9 — Key Formulas

Topic	Formula
Heron's Formula	A = √[s(s−a)(s−b)(s−c)] ; s = (a+b+c)/2
Surface Area of Sphere	TSA = 4πr²
Volume of Sphere	V = (4/3)πr³
CSA of Cone	CSA = πrl ; l = √(r² + h²)
TSA of Cone	TSA = πr(l + r)
Volume of Cone	V = ⅓πr²h
Hemisphere CSA	CSA = 2πr²
Hemisphere TSA	TSA = 3πr²
Hemisphere Vol	V = (2/3)πr³
Mean (direct)	x̄ = Σfᵢxᵢ / Σfᵢ
Mean (assumed)	x̄ = A + Σfᵢdᵢ / Σfᵢ
Median (ungrouped odd)	M = value at (n+1)/2
Median (ungrouped even)	M = avg of values at n/2, (n/2)+1
Polynomial Remainder	If P(a) = 0, then (x−a) is a factor
Distance Formula	d = √[(x₂−x₁)² + (y₂−y₁)²)
Midpoint Formula	M = ((x₁+x₂)/2, (y₁+y₂)/2)
Section Formula	P = ((mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n))
Area of Triangle	A = ½\|x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)\|
Logarithm	If aᵐ = N, then logₐ N = m
log product	log(M×N) = log M + log N
log quotient	log(M/N) = log M − log N

Class 10 — Key Formulas

Topic	Formula
AP nth term	aₙ = a + (n − 1)d
AP sum of n terms	Sₙ = n/2[2a + (n−1)d] = n/2(a + l)
Quadratic Formula	x = (−b ± √(b²−4ac)) / 2a
Discriminant	D = b² − 4ac
Sum of roots (α+β)	−b/a
Product of roots (αβ)	c/a
Distance Formula	d = √[(x₂−x₁)² + (y₂−y₁)²
Section Formula (int)	((mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n))
Section Formula (ext)	((mx₂−nx₁)/(m−n), (my₂−ny₁)/(m−n))
Centroid	((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3)
Area of Triangle (coord)	½\|x₁(y₂−y₃)+x₂(y₃−y₁)+x₃(y₁−y₂)\|
sin²θ + cos²θ	= 1
1 + tan²θ	= sec²θ
1 + cot²θ	= cosec²θ
sin(90°−θ)	= cos θ
tan(90°−θ)	= cot θ
Sector area	A = (θ/360°) × πr²
Arc length	l = (θ/360°) × 2πr
Segment area	(θ/360°)πr² − ½r²sinθ
Mean (step deviation)	x̄ = A + h(Σfᵢuᵢ/Σfᵢ)
Median (grouped)	l + [(n/2−cf)/f] × h
Mode (grouped)	l + [(f₁−f₀)/(2f₁−f₀−f₂)] × h
Probability	P(E) = favourable / total
Complement	P(not E) = 1 − P(E)

Text

─── Important Constants & Values ───

  π (pi)           ≈ 3.14159 ≈ 22/7
  √2               ≈ 1.414
  √3               ≈ 1.732
  √5               ≈ 2.236
  √10              ≈ 3.162
  e                ≈ 2.718
  Golden Ratio (φ)  ≈ 1.618

─── Square Roots to Remember ───
  √1=1  √4=2  √9=3  √16=4  √25=5  √36=6
  √49=7 √64=8 √81=9 √100=10 √121=11 √144=12
  √169=13 √196=14 √225=15 √256=16 √400=20
  √625=25 √900=30 √10000=100

─── Cube Roots to Remember ───
  ∛1=1  ∛8=2  ∛27=3  ∛64=4  ∛125=5
  ∛216=6 ∛343=7 ∛512=8 ∛729=9 ∛1000=10

─── Powers of 2 ───
  2¹=2  2²=4  2³=8  2⁴=16  2⁵=32  2⁶=64
  2⁷=128 2⁸=256 2⁹=512 2¹⁰=1024

─── Powers of 3 ───
  3¹=3  3²=9  3³=27 3⁴=81 3⁵=243

─── First 10 Prime Numbers ───
  2, 3, 5, 7, 11, 13, 17, 19, 23, 29

Text

─── Quick Unit Conversions ───

Length:
  1 km = 1000 m = 100000 cm = 1000000 mm
  1 m = 100 cm = 1000 mm
  1 mile = 1.609 km
  1 inch = 2.54 cm
  1 foot = 30.48 cm = 12 inches

Area:
  1 hectare = 10000 m² = 2.471 acres
  1 acre = 4047 m²
  1 m² = 10000 cm²

Volume:
  1 litre = 1000 cm³ = 1000 ml
  1 m³ = 1000 litres = 10⁶ cm³

Weight:
  1 kg = 1000 g
  1 tonne = 1000 kg

Time:
  1 hour = 60 min = 3600 s
  1 day = 24 hours
  1 year = 365 days = 52 weeks

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Ultimate Exam Tip: Before the exam, write down all formulas on one page as a quick reference. Focus on Class 10 formulas for board exams. For competitive exams, also memorize squares (1-30), cubes (1-10), and prime numbers up to 100. Always verify dimensional consistency — if area should be in m², check that your formula gives m².

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Common Pitfalls: (1) Forgetting units in the final answer. (2) Using diameter instead of radius in circle formulas. (3) Not converting all measurements to the same unit before calculating. (4) Using TSA when CSA is asked (and vice versa). (5) Forgetting that √(x²) = |x|, not x.

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